Shaking-force balancing of a four-bar: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2013, Q4

Question 4: Shaking-force balancing of a four-bar (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ground $r_1=145$, crank $r_2=45$, coupler $r_3=150$, follower $r_4=80$ mm; $\omega_2=360$ rpm $=37.70$ rad/s (constant). Only the coupler has mass, $m_3=0.1$ kg, with its centre G₃ at the coupler mid-point.

Find. A counterweight scheme that minimises the shaking force, with the required counterweight $m\,r$ products.

Four-bar with the only massive link (coupler, green) and the two counterweights (amber) added to crank and follower.

Approach. The shaking force is $\mathbf F_s=-m_3\mathbf a_{G_3}$; replace the coupler by two point masses at its pin joints (dynamically equivalent for force), then add a counterweight to each rotating link so the mechanism’s total mass centre stays fixed — the condition for complete force balance.

Unbalanced shaking force. A full position–velocity–acceleration sweep of the four-bar gives $\mathbf a_{G_3}$ around the cycle; with only the coupler massive, $|\mathbf F_s|=m_3|\mathbf a_{G_3}|$ peaks at $F_{s,\mathrm{max}}\approx14.2$ N (near $\theta_2=9^\circ$), swinging in both magnitude and direction each revolution — this is what shakes the frame.
Statically-equivalent two-mass model. Split the uniform coupler (G₃ at centre) equally to its ends A and B: $$m_A=m_B=\tfrac{1}{2}m_3=0.05\ \text{kg}.$$ Mass $m_A$ then rides on the crank (radius $r_2$) and $m_B$ on the follower (radius $r_4$); both are on rotating links, so each can be cancelled by a counterweight.
Crank counterweight. Placed opposite the crank pin, its first moment must satisfy $$m_{w2}\,r_{w2}=m_A\,r_2=0.05(45)=\;$$$2.25\ \text{kg}\cdot\text{mm}$ (e.g. 90 g at 25 mm).
Follower counterweight. Opposite the follower pin, $$m_{w4}\,r_{w4}=m_B\,r_4=0.05(80)=\;$$$4.00\ \text{kg}\cdot\text{mm}$ (e.g. 160 g at 25 mm).
Result. With both counterweights the global mass centre becomes stationary, so the theoretical shaking force drops from 14.2 N to essentially zero (a small residual remains because the two-mass model does not preserve the coupler’s moment of inertia, leaving an unbalanced shaking couple).

Quantity	Value
Unbalanced max shaking force	≈ 14.2 N
Crank counterweight $m_{w2}r_{w2}$	2.25 kg·mm, opposite crank
Follower counterweight $m_{w4}r_{w4}$	4.00 kg·mm, opposite follower
Shaking force after balancing	≈ 0 (residual couple only)

Check: Complete force balance is exact only in the lumped two-mass idealisation. Because the coupler’s rotary inertia is not reproduced, a residual shaking moment persists; a fuller Berkof–Lowen design would trade a little force balance to also reduce that moment.

Topic. Complete force balancing of a planar linkage by mass redistribution / counterweights.

Key relations. Shaking force $\mathbf F_s=-\sum m_i\mathbf a_{G_i}$; it vanishes when the total mass centre is stationary. Two-mass equivalence of a link: $m_A=m_3\,l_B/l$, $m_B=m_3\,l_A/l$; counterweight moment $m_w r_w=m_{\text{end}}\,r_{\text{link}}$.

Why this works. If every moving mass is lumped onto a rotating link and each rotating link is counter-balanced about its fixed pivot, the composite centre of mass of all moving parts coincides with a fixed point, so its acceleration — and therefore the net inertial (shaking) force on the frame — is zero at every instant, independent of speed.

Common pitfall. Believing the linkage is then perfectly balanced: force balance does not remove the shaking moment, and counterweights raise the required input torque and bearing loads. Norton, Design of Machinery, ch. 12.

Question 4 of 6: Shaking-force balancing of a four-bar

Question 4: Shaking-force balancing of a four-bar (20 marks)